You have a beaker containing 100ml of (slightly salty) water, standing on a digital scale. The scale has been calibrated to the beaker so that it shows only the weight of the water, which is exactly 100.0 grams. The beaker is perfectly cylindrical, and the depth of the water in it is exactly 5 cm.
You have a pencil which is half the density of the water. The rear end of the pencil is flat and has a surface area of 0.5 square centimetres. You put the rear end of the pencil into the water until it barely touches the bottom, and hold it here.
What will the scale say?
(Hand holding the pencil not pictured)
Answer
Well it seems to me that
The delta weight will be equal to the force of buoyancy against the pencil. This is Archimedes Principle in that the buoyant force is equal to the weight of the water displaced.
Also, as @Daniel Mathias points out
the water level goes up. So we have initially $5 \pi r^2 = 100$. Afterwards our new height is $x$ so $x \pi r^2 - 0.5 x = 100$. Solving this gives $x = 5.128205$.
Therefore
The weight of the water displaced is .5 sq. cm x 5.13 cm x 1000 kg/ cubic meter (the density of water) x 1/1000000 cubic meter/ cubic cm which comes out to be .002564 Kg or 2.564 grams
So the scale would read very close to
102.564 grams
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